<p>Does anyone know specifically how the exam is scored? I’ve heard that 115-180 once qualified for a 5, but the questions are easier and the scale is harsher since 2004 so that 123-180 is now required for a 5. Please confirm.</p>
<p>check this out</p>
<p><a href=“http://talk.collegeconfidential.com/ap-tests-preparation/499244-thread-ap-exam-curves.html[/url]”>http://talk.collegeconfidential.com/ap-tests-preparation/499244-thread-ap-exam-curves.html</a></p>
<p>I am very iffy on the stuff before it and remember, I am only taking Physics as an honors class, so I don’t know how much I’m missing compared to AP. Are there no cram sheets online for this, haha? I’ve been searching for like 30 minutes.</p>
<p>Ok, i just got a raw score of a 37.5 on the 2004 multiple choice. That thread says to multiply by 1.2857. That gives me a mc score of 48. so it also says to get a 4( whcih is what i want to get) i need 43 raw points. that comes out to 7 points per question without being multiplied by 1.1250. I think thats right, which seems pretty doable. can someone check my logic?</p>
<p>strikeb4ack
go on collegeboard and click on any free response and memorize the formula sheet they have in the beginning, IT WILL help for MC.
and rmr do as many problems as you cannnnn. practice makes perfect ya dig?</p>
<p>joem4d i dnt get wat u mean? you have to add ur fr score, dont ever assume you will get all of them right</p>
<p>does anyone know the answer to this?</p>
<p>An empty sled of mass K moves without friction across a frozen pond at speed vo. Two objects are
dropped vertically into the sled one at a time: first an object of mass m and then an object of mass 2m.
Later, the sled moves with speed vf . What would the final speed of the sled be if the objects were
dropped into it in reverse order?</p>
<p>vf, i just did that question</p>
<p>yeah, the answer is vf. You can do this by doing .5k(vo^2)= .5(m +2m + K)(vf^2). If the two masses were dropped in different orders the last equation would be .5(2m +m + K)(vf^2) which is the same thing. Also, is angular momentum and velocity in the physics b test because the barron’s book goes over it, but college board says that it is only on the physics c test?</p>
<p>no its not part of the ap exam
princeton and barrons always do that
that tells you something about them</p>
<p>angular v and p are not on PR.</p>
<p>Wrong. angular p is on PR. it goes off by saying
“angular momentum, L = mvr” on the top of page 352</p>
<p>Angular momentum may be in PR, but the only thing you have to know for the AP Physics B test is that the circumference of an electron’s orbit must be equal to a whole number of wavelengths for an atom’s stability.</p>
<p>I’ve actually found out in the long run that if you want to get a perfect on your multiple choice, you pretty much need to memorize the formula sheet.</p>
<p>I have a question, but it requires a diagram and I don’t have a scanner or anything, so I’m just gonna do my best to describe it…bear with me XD</p>
<p>Okay so there’s 2 masses, m1 and m2. The two are connected by a light string, with m1 hanging off the edge of the table by a pulley. m2 stays on the smooth frictionless table. The question asks what the acceleration of m2 is. I always thought that the acceleration of the masses in this type of question is always [(m1-m2)/(m1+m2)]g, but the answer is [m1/(m1+m2)]g, apparently…can anyone please explain why?</p>
<p>Thanks!</p>
<p>you just have to draw the force body diagrams and it wouldnt take long to figure out… its just because the acceleration of the block m2 has nothing to do with gravity and is just produced by the tension in the string…
so if a is the acceleration:
m1g-T=m1a
T=m2a
a=m1*g/(m1+m2)</p>
<p>oceanangel//it is a bad habit to memorize rules like that and apply it to a seemingly similar situation. I believe ur equation is actually for pulleys with two ends of the spring, and each is attached to masses that are hanging VERTICALLY.
This case asks for one that stays horizontally and one that is hanging. newton’s second law. for m2 that is on a horizontal surface, the weight and normal forces cancel out so there is no net force that is acting upon the direction of interest.
sigma F,y = m1 * g = (m1 + m2) * a
Therefore a = m1g/(m1 + m2)</p>
<p>Now i will briefly show my ASCII artistic skills, although i consider myself to be a freaking novice.</p>
<p>m2---------------------------------O
===========================.||
<strong><em>.TABLE</em></strong>_||
AAAAAAAAAAAAAAAAAAAAAAAAAAA.||
AAAAAAAAAAAAAAAAAAAAAAAAAAA.||
AAAAAAAAAAAAAAAAAAAAAAAAAAA.|m1
AAAAAAAAAAAAAAAAAAAAAAAAAAA.||
AAAAAAAAAAAAAAAAAAAAAAAAAAA.||
AAAAAAAAAAAAAAAAAAAAAAAAAAA.|v
AAAAAAAAAAAAAAAAAAAAAAAAAAA.|F1 = m1g</p>
<p>Here’s another way of thinking of it:</p>
<p>The acceleration of m1 is the same as the acceleration of m2. The force of m1 would normally be (m1)g, which would be used to solve for the acceleration. However, m1 is “pulling” both itself AND m2. So, the tension is in reality (m1+m2)a, which is equal to (m1)g. So, set (m1+m2)a=(m1)g and solve for a.</p>
<p>What is everyone doing for their last minute studying?</p>
<p>Just looking over my notes and PR practice exams</p>
<p>Okay, thank you everyone for your help^^ I understand now hehe</p>
<p>me too, Sagert, im lookin at my notes + crammin formula sheet into my head right now lol.</p>